Question

4. Show that the pair of lines 6x^2 – 5xy – 6y² = 0 , 6x’ – 5xy – 6y^2 +x+5y-1=0.
forms a square.​

Answer 1

There are two equations of a pair of straight lines so there are a total of 4 lines. To show that they form a square, the following should be verified.

• There should be two pairs of parallel lines among the four lines...(1)

• There should be two pairs of perpendicular lines among the four lines...(2)

• The distance between the parallel lines should be same...(3)

Let us find out the equations of individual lines. Consider,

$\small\longrightarrow 6x^2-5xy-6y^2=0$

Dividing by y²,

$\small\longrightarrow 6\left(\dfrac{x}{y}\right)^2-5\left(\dfrac{x}{y}\right)-6=0$

Solving by quadratic equation formula,

$\small\text{ \longrightarrow\dfrac{x}{y}=\dfrac{5\pm\sqrt{(-5)^2-4\cdot6\cdot(-6)}}{2\cdot6} }$

$\small\longrightarrow\dfrac{x}{y}=\dfrac{3}{2}\quad OR\quad \dfrac{x}{y}=-\dfrac{2}{3}$

By cross multiplication,

$\small\longrightarrow2x-3y=0\quad OR\quad 3x+2y=0$

These are the individual lines represented by the equation.

$\small\longrightarrow 6x^2-5xy-6y^2=(2x-3y)(3x+2y)$

For 6x² - 5xy - 6y² + x + 5y - 1 = 0, assume,

$\small\longrightarrow 6x^2-5xy-6y^2+x+5y-1=(2x-3y+m)(3x+2y+n)$

$\begin{aligned}\longrightarrow\ \ &\small(2x-3y)(3x+2y)+x+5y-1\\=\ \ &\small(2x-3y)(3x+2y)+m(3x+2y)+n(2x-3y)+mn\end{aligned}$

Cancelling like terms,

$\small\longrightarrow x+5y-1=m(3x+2y)+n(2x-3y)+mn$

$\small\longrightarrow x+5y-1=(3m+2n)x+(2m-3n)y+mn$

Equating corresponding coefficients,

• $\small3m+2n=1$

• $\small2m-3n=5$

Solving them we get,

• $\smallm=1$

• $\smalln=-1$

Therefore,

$\small\longrightarrow 6x^2-5xy-6y^2+x+5y-1=(2x-3y+1)(3x+2y-1)$

So the four lines are,

• $\smallL_1:2x-3y=0$

• $\smallL_2:3x+2y=0$

• $\smallL_3:2x-3y+1=0$

• $\smallL_4:3x+2y-1=0$

Slopes of L₁ and L₃ are equal (2/3) and those of L₂ and L₄ are equal (-3/2) [Recall that slope of the line ax + by + c = 0 is -a/b]. So L₁ ║ L₃ and L₂ ║ L₄. Hence there are two pairs of parallel lines and (1) is true.

Here L₁ ⊥ L₂ and L₃ ⊥ L₄ as product of the slopes equal -1. Hence there are two pairs of perpendicular lines and (2) is true.

Now distance between the parallel lines L₁ and L₃ is,

$\small\text{ \longrightarrow d_{13}=\dfrac{|1-0|}{\sqrt{2^2+(-3)^2}}=\dfrac{1}{\sqrt{13}} }$

and distance between the parallel lines L₂ and L₄ is,

$\small\text{ \longrightarrow d_{24}=\dfrac{|1-0|}{\sqrt{2^2+(-3)^2}}=\dfrac{1}{\sqrt{13}} }$

The distances are same, so (3) is true.

Hence the lines form a square as the three conditions are satisfied.